Practice 20 Simplification/ Approximation Important Questions with Answers for Prelims Exam

IBPS Clerk 2021 Numerical Ability Solved Mock Paper: Get Important Simplification/ Approximation Questions with Answers to score high marks in Numerical Ability section of IBPS Clerk Prelims Exam to be held on 12th, 18th & 19th December 2021 in online mode.

IBPS Clerk 2021 Numerical Ability Solved Mock Paper: The Institute for Banking Personnel Selection (IBPS) is going to hold Preliminary Online Exams for the recruitment of 7858 vacancies of candidates in the clerical cadre of public sector banks under CRP Clerks-XI drive on 12^th, 18^th & 19^th December 2021. The numerical Ability section is one of the deciding areas for a candidate in the merit list of the IBPS Clerk Exam. In this article, we are going to share the Important Simplification/ Approximation Questions to score high marks in IBPS Clerk Prelims 2021 Exam. Let’s first look at the IBPS CLERK Prelims 2021 Exam Pattern in brief:

IBPS Clerk 2021 Prelims Exam Pattern

IBPS Clerk 2021 Exam will consist of 100 objective multiple-choice questions of 100 marks. The exam will have three sections (with separate timings for each section) as follows:

IBPS Clerk 2021 Prelims Exam: Numerical Ability (35 Marks)

Numerical Ability will test the candidates’ mathematical skills. The questions will be designed to test the ability of appropriate use of numbers and number sense of the candidate. Getting a good score in this section demands in-depth knowledge of all the formulas and the pattern of the question asked.

IBPS Clerk 2021 Prelims Exam 20 Important Simplification/ Approximation Questions with Answers

Let’s look at some important Simplification/ Approximation Questions with Answers to score high marks in the IBPS Clerk Prelims 2021 Exam.

Directions (1 – 20): What should come in place of question mark (?) in the following questions?

1. 621 – 23.829 – 9.302 – 2.214 = ?

1. 281
2. 272
3. 276
4. 254
5. None of these

Answer: (c)

Explanation: 78.621 – 23.829 – 9.302 – 2.214 = 43.276

2. ?% of 550 + 36% of 285 = 306.1

1. 34
2. 32
3. 21
4. 37
5. None of these

Answer: (d)

Explanation: x ÷ 100 X 550 + 36 ÷ 100 X 285 = 1.

306.1

x= 37

3. 64⁴ ÷ 8⁵

1. 4096
2. 512
3. 64
4. 8
5. None of these

Answer: (b)

Explanation: (8²)⁴/8⁵ = 512

4. 518 + 52 x 0.50 – 29 = ?

1. 625
2. 660
3. 640
4. 509
5. None of these

Answer: (d)

Explanation: 518 + 52 x 0.50 – 29 = 509

5. 4(16/17) X 1(11/16) ÷ (7/38) = ?

1. 12(3/17)
2. 45(9/34)
3. 12(21/34)
4. 36(8/17)
5. None of these

Answer: (b)

Explanation: 4(16/17) X 1(11/16) ÷ (7/38) = (84/17) X (27/16) X (38/7) = 1539/34 = 45(9/34)

6. 5 × 0.9 + 0.7 × 0.7 + 0.07 ÷ 7 – 0.07 ÷ 0.014 = ?

1. 20
2. 7
3. 3
4. 0
5. None of the above

Answer: d)

Explanation: 5 × 0.9 + 0.7 × 0.7 + 0.07 ÷ 7 – 0.07 ÷ 0.014 = 4.5 + 0.49 + 0.01 – 5 = 0

7. 25% of √1296 ÷ 0.5 = ? ⤬ 27

1. 1/3
2. 3
3. 2/3
4. 2
5. None of the above

Answer: c)

Explanation: Let the number be x.

√1296 = 36

25% of 36 = 9

9 ÷ 0.5 = 18

Thus, 27x = 18 = x = 2/3

8. 12² ⤬ 18 ÷ 24 + 193 + 12 ⤬ 5 = (?)²

1. 3√2
2. 4√2
3. 5√2
4. 19
5. None of the above

Answer: d)

Explanation: 12² ⤬ 18 ÷ 24 + 193 + 12 ⤬ 5 = 144 ⤬ (3/4) + 193 + 60

108 + 203 + 60 = 361 = (19)²

9. 58% of 1688 + 36% of 934 – 215.28 =?

1. 28
2. 28
3. 1000
4. 1100
5. None of the above

Answer: d)

Explanation: (58/100) × 1688 + (36/100) × 934 – 215.28

= 979.04 + 336.24 – 215.28

= 1315.28 – 215.28 = 1100

10. 55% of √3136 x 7 =? + 154

1. 61
2. 62
3. 6
4. 8
5. None of the above

Answer: c)

Explanation: (55/100) × 56 × 7 = (55/100) × 392 = 215.6

215.6 = ? + 154

? = 61.6

11. ³√68921 X (450 ÷ 11) = (?)²

1. 36
2. 35
3. 39
4. 41
5. None of these

Answer: (d)

Explanation: (?)² = ³√68921 X (450 ÷ 11) = (³√41 X 41 X 41) X 40.99 = 1681 = 41 X 41 = (41)²

12. 5% of 1480 + 63.002 X 9.995 – 61.899 = ?

1. 3500
2. 4400
3. 4108
4. 4650
5. 4760

Answer: (c)

Explanation: ? = (277.5 X 1480)/100 + 63 X 10 – 62 = 4107 + 63 – 62 = 4108

13. (7/16) X 6022.66 + (11/20) X 5822.44 = ?

1. 5837
2. 5290
3. 6310
4. 5300
5. 5929

Answer: (a)

Explanation: ? = (7/16) X 6022.66 + (11/20) X 5822.44

= 2635.06 + 3202.1 ≈ 5837

14. 7975 ÷ 545 X ³√515 = ?

1. 140
2. 150
3. 148
4. 154
5. 132

Answer: (e)

Explanation: ? = (7975/545) X ³√728 (Therefore 515 ≈ 512)

= 14.63 X 9 = 131.67

15. 089 X 5.27 X 6.004 X 222.918 = ?

1. 2800
2. 2350
3. 2850
4. 2336
5. None of these

Answer: (d)

Explanation:

? = 190 X 5.25 + 6 X 223

= 997.5 + 1338 ≈ 2335.5

16. (57% of 1552 − 46% of 1512)/73= ?

1. 6
2. 6
3. 6
4. 4
5. None of the above

Answer: c)

Explanation: (57% of 1552 – 46% of 1512)/73

= 884.64 – 695.52/73 = 2.6

17. (22.13)² + (13.96)² – (7.11)² = ?

1. 644
2. 564
3. 784
4. 634
5. None of the above

Answer: d)

Explanation: (22.13)² + (13.96)² – (7.11)²

= 490 + 195 – 51 = 634

18. 26% of 900 × ? % of 1042 = 4168

1. 7
2. 7
3. 7
4. 7
5. None of the above

Answer: b)

Explanation: 26% of 900 X x% of 1042 = 4168

234 X x% of 1042 = 4168

x= (4168 × 100)/(234 × 1042) = 1.7

19. (0.16)⁷/(0.064)³ × (0.4)² = (0.4)^?

1. 5
2. 6
3. 7
4. 8
5. None of the above

Answer: c)

Explanation: (0.16)⁷/(0.064)³ × (0.4)²

= (0.4)¹⁴/(0.4)⁹ × (0.4)²

= (0.4)¹⁶/(0.4)⁹ = (0.4)⁷

20. 5 + 5.55 + 0.5 + 55.05 + 555 = ?

1. 611
2. 621
3. 600
4. 651
5. None of the above

Answer: b)

Explanation: 5 + 5.55 + 0.5 + 55.05 + 555 = 621.1

Remember that there are separate timings for each section. There will be a sectional cut-off. Candidates who obtain cut-off marks in the Prelims exam get shortlisted for the Mains exam and those who obtain cut-off marks in the Mains exam get selected for the Provisional Allotment. However, for the final merit list, marks obtained by candidates in the Mains exam will only be considered. So your job is to simply maximize your score however you can. So, try to maintain your speed and accuracy while solving the paper during the examination.